1000 Solved Problems in Modern Physics

(Grace) #1

1.3 Solutions 77


The volume:

V=π

∫a

0

y^2 dx

Therefore, dropping off the constant factors
K=y^2 +λy(1+y′^2 )^1 /^2
which must satisfy the Euler’s equation
∂K
∂x


d
dx

(K−y′

∂K

∂y′

)= 0

It is convenient to use the above form asKdoes not explicitly containx, and
∂K
∂x=0. Therefore,

K−y′

∂K

∂y′

=y^2 +λy(1+λy′^2 )

(^12)
−λyy′^2 (1+y′^2 )−
(^12)
= 0
Nowy=0atx=0 and atx=awhich can be true ifC=0. Hence
y^2 +λy(1+y′^2 )−^1 /^2 = 0
Ory=−λ(1+y′^2 )−^1 /^2
Solving fory′,
dy
dx


=

1

y

(λ^2 −y^2 )^1 /^2

Integrating,
−(λ^2 −y^2 )

1

(^2) =x−x 0
Or (x−x 0 )^2 +y^2 =λ^2
This is the equation to a sphere with the centre on thex-axis atx 0 , and of
radiusλ.


1.3.13 StatisticalDistribution..............................


1.93 (a)

∑∞

x= 0
Px=

∑∞

x= 0

e−mmx
x!

=e−m

(

1 +

m
1!

+

m^2
2!

+···

)

=e−m×e+m= 1
Thus the distribution is normalized.
(b)<x>=

∑∞

x= 0
xPx=

∑∞

x= 0

xe−mmx
x!

=

∑∞

x= 0

e−mmx
(x−1)!

=e−m

(

m+

m^2
1!

+

m^3
2!

+···

)

(∵(−1)!=∞)

=me−m

(

1 +

m
1!

+

m^2
2!

+···

)

=me−m×em=m
Free download pdf